Question

How do you find an equation for the horizontal tangent plane to the surface $z=4{(x-1)}^2+3{(y+1)}^2$ ?

Answer 1

Hint: In order to find a solution for this question, we have to remember that a horizontal plane is tangent to a curve in the space in its points of maximum, minimum or saddle. Tangent planes to a surface are planes that just touch the surface at the point and are “parallel” to the surface at the point.

Complete step by step solution:
As we have our equation as:
$z=4{(x-1)}^2+3{(y+1)}^2$
We will first have to calculate the two partial derivative,
For $x$ , since we have: $4{(x-1)}^2$ so we get:
$\Rightarrow {\displaystyle \frac{\partial z}{\partial x}}=4\times 2(x-1)$
For $y$ , since we have: $3{(y+1)}^2$ so we get:
$\Rightarrow {\displaystyle \frac{\partial z}{\partial y}}=3\times 2(y+1)$
$\Rightarrow {\displaystyle \frac{\partial z}{\partial y}}=6(y+1)$
Now put $=0$ in the system of two previous equations (that is in place of derivative) and thus it will give us the stationary points:
$8(x-1)=0$
$6(y+1)=0$
And therefore we get:
$\begin{array}{rl}& \Rightarrow x=1\\ & \Rightarrow y=-1\end{array}$
From this we can conclude that $z=0$
Now we have to find what type of stationary point $$P(1,-1,0)$$ is.
So, to do that we have to calculate the four partial derivative of the second order:
$\Rightarrow {\displaystyle \frac{\partial z}{\partial x}}=8(x-1)$
By applying second order of above derivative, we get:
$\Rightarrow {\displaystyle \frac{{\partial }^{2}z}{\partial x^2}}=8$
Also, for:
$\Rightarrow {\displaystyle \frac{\partial z}{\partial y}}=6(y+1)$
By applying second order of above derivative, we get:
$\Rightarrow {\displaystyle \frac{{\partial }^{2}z}{\partial y^2}}=6$
With this we can conclude as:
${\displaystyle \frac{{\partial }^{2}z}{\partial x\partial y}}={\displaystyle \frac{{\partial }^{2}z}{\partial y\partial x}}=0$
Also, the hessian matrix will be:
$\left[\begin{array}{cc}8& 0\\ 0& 6\end{array}\right]$
Since its determinant in P is positive $48$ and ${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}}$ is positive $8$
Therefore, the point P is a local minimum, so there exists a horizontal tangent plane.
Also, since it has to pass from P which $$z=0$$, its equation is $$z=0$$

Note: If the tangent plane is horizontal, the gradient must pointed in the $z$-direction, therefore the $x$ and $y$ components are $0$. So it follows $x=1,y=-1$. Therefore, this gives us a point that is on the plane. Since the tangent plane and the surface touch at $(x_0,y_0)$the following point will be on both the surface and the plane $$(x_0,y_0,z_0)=(x_0,y_0,f(x_0,y_0))$$.
Alternate method:
If we can recognize the curve of an equation:
As we can see this function is the equation of an elliptic paraboloid with concavity upwards.
Also, Since $z$ is surely positive or zero because the sum of two quantities will be positive or zero.
Hence the horizontal plane is tangent to a curve in the space in its points is minimum, then the vertex, is zero, and this happens where:
$$x=1$$ and $$y=-1$$
So the vertex is $$V(1,-1,0)$$and therefore the plane requested is the floor of the $3$-dimensional space:
Therefore,
$z=0$ is the answer.